Question

1. Normal Approximation to Binomial Assume

n = 10, p = 0.1.

a. Use the Binomial Probability function to compute the P(X = 2)

b. Use the Normal Probability distribution to approximate the P(X = 2)

c. Are the answers the same? If not, why?

Answer #1

X ~ B ( n = 10 , P = 0.1 )

Part a)

part b)

Using Normal Approximation to Binomial

Mean = n * P = ( 10 * 0.1 ) = 1

Variance = n * P * Q = ( 10 * 0.1 * 0.9 ) = 0.9

Standard deviation = √(variance) = √(0.9) = 0.9487

P ( X = 2 )

Using continuity correction

P ( n - 0.5 < X < n + 0.5 ) = P ( 2 - 0.5 < X < 2 + 0.5
) = P ( 1.5 < X < 2.5 )

X ~ N ( µ = 1 , σ = 0.9487 )

P ( 1.5 < X < 2.5 )

Standardizing the value

Z = ( X - µ ) / σ

Z = ( 1.5 - 1 ) / 0.9487

Z = 0.527

Z = ( 2.5 - 1 ) / 0.9487

Z = 1.5811

P ( 0.53 < Z < 1.58 )

P ( 1.5 < X < 2.5 ) = P ( Z < 1.58 ) - P ( Z < 0.53
)

P ( 1.5 < X < 2.5 ) = 0.9431 - 0.7009

P ( 1.5 < X < 2.5 ) = 0.2422

Part c)

Answers are not same because, Binomial distribution is discrete and Normal distribution is continuous distribution.

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